Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 1 - Functions and Limits - Review - Exercises: 30

Answer

$\dfrac{1}{3}$

Work Step by Step

$\lim\limits_{t \to 2}\dfrac{t^2-4}{t^3-8}$ Factor: $\lim\limits_{t \to 2}\dfrac{(t-2)(t+2)}{(t-2)(t^2+2t+4)}$ Simplify: $\lim\limits_{t \to 2}\dfrac{t+2}{t^2+2t+4}$ Plug in 2: $\dfrac{2+2}{2^2+2(2)+4}=\dfrac{4}{12}=\dfrac{1}{3}$ Therefore, $\lim\limits_{t \to 2}\dfrac{t^2-4}{t^3-8}=\dfrac{1}{3}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.